Q106 / 200 · NEET 2011 (Screening) · Physics
The instantaneous angular position of a point on a rotating wheel is given by the equation $Q(t)=2 t^3-6 t^2$
The torque on the wheel becomes zero at
- A$t=0.5 \mathrm{~s}$
- B$t=0.25 \mathrm{~s}$
- C$t=2 \mathrm{~s}$
- Correct answerD$t=1 \mathrm{~s}$
Solution
According to question, torque, $\tau=0$
Its means that, $\alpha=0$
$$
\alpha=\frac{d^2 \theta}{d t^2}
$$
Given
So
$$
\theta(t)=2 t^3-6 t^2
$$
$$
\begin{aligned}
\frac{d \theta}{d t} & =6 t^2-12 t \\
\frac{d^2 \theta}{d t^2} & =12 t-12 \quad\left(\because \alpha=\frac{d^2 \theta}{d t^2}=0\right) \\
12 t-12 & =0 \\
t & =1 \mathrm{~s}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.